ℝ2∖C is path connected if C is countable

Theorem 1.

We use ℝ2 simply as an example; an analogous proof will work for any ℝn,n>1.

Proof.

Fix a point P not in C. The strategy of the proof is to construct a path px from any x∈ℝ2∖C to P. If we can do this then for any d,d′∈ℝ2∖C we may construct a path from d to d′ by first following pd and then following pd′ in reverse.

Fix x∈ℝ2∖C, and consider the set of all (straight) lines through x. There are uncountably many of these and they meet in the single point x, so not all of them contain a point of C. Choose one that doesn’t and move along it: your distance from P takes on uncountably many values, and hence at some point this distance r from P is not shared by any point of C. The whole of the circle with radius r, centre P, lies in ℝ2∖C so we may move around it freely.

Consider all lines through P: these all intersect this circle, and there are uncountably many of them so we may choose one, say L, that contains no point of C. Moving around the circle until we meet L and then following it inwards completes our path form x to P.
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Corollary 1.

Proof.

Suppose that f-1⁢(0) is countable. ℝ2 can be written as the disjoint union

f-1⁢(0)∪f-1⁢((-∞,0))∪f-1⁢((0,∞))

where the last two sets are open (as f is continuous), non-empty (as f is onto) and disjoint. Since pathwise connected is the same as connected for Hausdorff spaces, we have that ℝ2∖f-1⁢(0) is not path connected, contradicting the theorem.
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